• Coming Out as a Gay Orthodox Talmud Teacher

It would have been even better if it had been in *The Guardian*, but this is an imperfect world.

• Queen and Not Heard – *The African Queen*, C.S. Forester (1935)

• Ley Ho, Let’s Go! – *Britannia Obscura: Mapping Britain’s Hidden Landscapes*, Joanne Parker (Vintage 2014)

• Hymn Pickings – *The Church Hymnary*, Various (Oxford University Press 1927)

• Clock, Stock and Biswell – *A Clockwork Orange: The Restored Edition*, Anthony Burgess, edited by Andrew Biswell (Heinemann 2012)

• Vid Kid – *Violated by Video: The Eldritch Chills of an ’Eighties Childhood from Video Nasties to Nazi Xploitation*, Paolo Nanderson (TransVisceral Books 2018)

Or Read a Review at Random: RaRaR

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Cats are of divers colours, but for the most part griseld, like to congealed ise, which cometh from the condition of her meat: her head is like unto the head of a Lion, except in her sharp ears: her flesh is soft and smooth: her eyes glister above measure, especially when a man cometh to see them on the suddain, and in the night they can hardly be endured, for their flaming aspect. Wherefore Democritus describing the Persian Smaragde saith that it is not transparent, but filleth the eye with pleasant brightness, such as is in the eyes of Panthers and Cats, for they cast forth beams in the shadow and darkness, but in sunshine they have no such clearness, and thereof Alexander Aphrodise giveth this reason, both for the sight of Cats and Bats, that they have by nature a most sharpe spirit of seeing. — Edward Topsell, *Historie of Foure-Footed Beastes* (1658).

• If the die rolls *n*, then player #*n* wins.

Now try a different game with six players and one die. Player #1 rolls the die. If he gets 1, he wins the game. If not, then he leaves the game and player #2 rolls the die. If he gets 2, he wins the game. If not, then he leaves the game and player #3 rolls the die. And so on. You could call this a sequential game, because the players are playing in sequence. It has two rules:

• If player #*n* rolls *n* on the die, then he wins.

• If player #*n* doesn’t roll *n*, then player *n*+1 rolls the die.

Is it a fair game? No, definitely not. Player #1 has the best chance of winning. 1/6 or 16.6% of the time he rolls 1 and wins the game. 5/6 of the time, he rolls 2, 3, 4, 5 or 6 and passes the die to player #2. Now player #2 has a 1/6 chance of rolling a 2 and winning. But he has the opportunity to roll the die only 5/6 of the time, so his chance of winning the game is 1/6 * 5/6 = 5/36 = 13.8%. However, if player #2 rolls a 1, 3, 4, 5 or 6, then he loses and player #3 rolls the die. But player #3 has that opportunity only 5/6 * 5/6 = 25/36 of the time. So his chance of winning is 1/6 * 25/36 = 11.57%. And so on.

To put it another way, if the six players play 46656 = 6^6 games under the sequential rules, then on average:

• Player #1 wins 7776 games

• Player #2 wins 6480 games

• Player #3 wins 5400 games

• Player #4 wins 4500 games

• Player #5 wins 3750 games

• Player #6 wins 3125 games

• 15625 games end without a winner.

In other words, player #1 is 20% more likely to win than player #2, 44% more likely than player #3, 72.8% more likely than player #4, 107% more likely than player #5, and 148.8% more likely than player #6. Furthermore, player #2 is 20% more likely to win than player #3, 44% more likely than player #4, 72.8% more likely than player #5, and so on.

But there is a simple way to make the sequential game perfectly fair, so long as it’s played with a fair die. At least, I’ve thought of a simple way, but there might be more than one.

To make the sequential game fair, you add an extra rule:

1. If player #*n* rolls *n* on the die, he wins the game.

2. If player #*n* rolls a number greater than *n*, he loses and the die passes to player *n*+1.

3. If player #*n* rolls a number less than *n*, then he rolls again.

Let’s run through a possible game to see that it’s fair. Player #1 rolls first. He has a 1/6 chance of rolling a 1 and winning the game. However, 5/6 of the time he loses and passes the die to player #2. If player #2 rolls a 1, he rolls again. In other words, player #2 is effectively playing with a five-sided die, because all rolls of 1 are ignored. Therefore, he has a 1/5 chance of winning the game at that stage.

But hold on: a 1/5 chance of winning is better than a 1/6 chance, which is what player #1 had. So how is the game fair? Well, note the qualifying phrase at the end of the previous paragraph: *at that stage*. The game doesn’t always reach that stage, because if player #1 rolls a 1, the game is over. Player #2 rolls only if player doesn’t roll 1, which is 5/6 of the time. Therefore player #2’s chance of winning is really 1/5 * 5/6 = 5/30 = 1/6.

However, 4/5 of the time player #2 rolls a 3, 4, 5 or 6 and the die passes to player #3. If player #3 rolls a 1 or 2, he rolls again. In other words, player #3 is effectively playing with a four-sided die, because all rolls of 1 and 2 are ignored. Therefore, he has a 1/4 chance of winning the game at that stage.

A 1/4 chance of winning is better than a 1/5 chance and a 1/6 chance, but the same reasoning applies as before. Player #3 rolls the die only 5/6 * 4/5 = 20/30 = 2/3 of the time, so his chance of winning is really 1/4 * 2/3 = 2/12 = 1/6.

However, 3/4 of the time player #2 rolls a 4, 5 or 6 and the die passes to player #4. If player #4 rolls a 1, 2 or 3, he rolls again. In other words, player #4 is effectively playing with a three-sided die, because all rolls of 1, 2 and 3 are ignored. Therefore, he has a 1/3 chance of winning the game at that stage. 1/3 > 1/4 > 1/5 > 1/6, but the same reasoning applies as before. Player #4 rolls the die only 5/6 * 4/5 * 3/4 = 60/120 = 1/2 of the time, so his chance of winning is really 1/3 * 1/2 = 1/6.

And so on. If the die reaches player #5 and he gets a 1, 2, 3 or 4, then he rolls again. He is effectively rolling with a two-sided die, so his chance of winning is 1/2 * 5/6 * 4/5 * 3/4 * 2/3 = 120/720 = 1/6. If player #5 rolls a 6, he loses and the die passes to player #6. But there’s no need for player #6 to roll the die, because he’s bound to win. He rolls again if he gets a 1, 2, 3, 4 or 5, so eventually he must get a 6 and win the game. If player #5 loses, then player #6 automatically wins.

It’s obvious that this form of the game will get slower as more players drop out, because later players will be rolling again more often. To speed the game up, you can refine the rules like this:

1. If Player #1 rolls a 1, he wins the game. Otherwise…

2. If player #2 rolls a 2, he wins the game. If he rolls a 1, he rolls again. Otherwise…

3. Player #3 rolls twice and adds his scores. If the total is 3, 4 or 5, he wins the game. Otherwise…

4. Player #4 rolls once. If he gets 1 or 2, he wins the game. Otherwise…

5. Player #5 rolls once. If he gets 1, 2 or 3, he wins the game. Otherwise…

6. Player #6 wins the game.

Only player #2 might have to roll more than twice. Player #3 has to roll twice because he needs a way to get a 1/4 chance of winning. If you roll two dice, there are:

• Two ways of getting a total of 3: roll #1 is 1 and roll #2 is 2, or vice versa.

• Three ways of getting a total of 4 = 1+3, 3+1, 2+2.

• Four ways of getting 5 = 1+4, 4+1, 2+3, 3+2.

This means player #3 has 2 + 3 + 4 = 9 ways of winning. But there are thirty-six ways of rolling one die twice. Therefore player #3 has a 9/36 = 1/4 chance of winning. Here are the thirty-six ways of rolling one die twice, with asterisks marking the winning totals for player #3:

01. (1,1)

02. (1,2)*

03. (2,1)*

04. (1,3)*

05. (3,1)*

06. (1,4)*

07. (4,1)*

08. (1,5)

09. (5,1)

10. (1,6)

11. (6,1)

12. (2,2)*

13. (2,3)*

14. (3,2)*

15. (2,4)

16. (4,2)

17. (2,5)

18. (5,2)

19. (2,6)

20. (6,2)

21. (3,3)

22. (3,4)

23. (4,3)

24. (3,5)

25. (5,3)

26. (3,6)

27. (6,3)

28. (4,4)

29. (4,5)

30. (5,4)

31. (4,6)

32. (6,4)

33. (5,5)

34. (5,6)

35. (6,5)

36. (6,6)

• They say, then, that Hipparchus the Pythagorean, being guilty of writing the tenets of Pythagoras in plain language, was expelled from the school, and a pillar raised for him as if he had been dead. — Clement of Alexandria, *The Stromata*, 2.5.9.57.3-4

• a

Then, if necessary, you reduce the numerator and denominator to their simplest possible terms. So the sequence starts like this:

• 0/1, 1/1

To create the next stage, find the mediant of the two fractions above: (0+1) / (1+1) = 1/2

• 0/1, 1/2, 1/1

For the next stage, there are two mediants to find: (0+1) / (1+2) = 1/3, (1+1) / (2+3) = 2/3

• 0/1, 1/3, 1/2, 2/3, 1/1

Note that 1/2 is the mediant of 1/3 and 2/3, that is, 1/2 = (1+2) / (3+3) = 3/6 = 1/2. The next stage is this:

• 0/1, 1/4, 1/3, 2/5, 1/2, 3/5, 2/3, 3/4, 1/1

Now 1/2 is the mediant of 2/5 and 3/5, that is, 1/2 = (2+3) / (5+5) = 5/10 = 1/2. Further stages go like this:

• 0/1, 1/5, 1/4, 2/7, 1/3, 3/8, 2/5, 3/7, 1/2, 4/7, 3/5, 5/8, 2/3, 5/7, 3/4, 4/5, 1/1

• 0/1, 1/6, 1/5, 2/9, 1/4, 3/11, 2/7, 3/10, 1/3, 4/11, 3/8, 5/13, 2/5, 5/12, 3/7, 4/9, 1/2, 5/9, 4/7, 7/12, 3/5, 8/13, 5/8, 7/11, 2/3, 7/10, 5/7, 8/11, 3/4, 7/9, 4/5, 5/6, 1/1

`• 0/1, 1/7, 1/6, 2/11, 1/5, 3/14, 2/9, 3/13, 1/4, 4/15, 3/11, 5/18, 2/7, 5/17, 3/10, 4/13, 1/3, 5/14, 4/11, 7/19, 3/8, 8/21, 5/13, 7/18, 2/5, 7/17, 5/12, 8/19, 3/7, 7/16, 4/9, 5/11, 1/2, 6/11, 5/9, 9/16, 4/7, 11/19, 7/12, 10/17, 3/5, 11/18, 8/13, 13/21, 5/8, 12/19, 7/11, 9/14, 2/3, 9/13, 7/10, 12/17, 5/7, 13/18, 8/11, 11/15, 3/4, 10/13, 7/9, 11/14, 4/5, 9/11, 5/6, 6/7, 1/1
`

The Farey sequence is actually a fractal, as you can see more easily when it’s represented as an image:

Farey fractal stage #1, representing 0/1, 1/2, 1/1

Farey fractal stage #2, representing 0/1, 1/3, 1/2, 2/3, 1/1

Farey fractal stage #3, representing 0/1, 1/4, 1/3, 2/5, 1/2, 3/5, 2/3, 3/4, 1/1

Farey fractal stage #4, representing 0/1, 1/5, 1/4, 2/7, 1/3, 3/8, 2/5, 3/7, 1/2, 4/7, 3/5, 5/8, 2/3, 5/7, 3/4, 4/5, 1/1

Farey fractal stage #5

Farey fractal stage #6

Farey fractal stage #7

Farey fractal stage #8

Farey fractal stage #9

Farey fractal stage #10

Farey fractal (animated)

That looks like the slope of a hill to me, so you could call it a Farey fract-hill. But Farey fract-hills or Farey fractals aren’t confined to the unit interval, 0/1 to 1/1. Here are Farey fractals for the intervals 0/1 to *n*/1, *n* = 1..10:

Farey fractal for interval 0/1 to 1/1

Farey fractal for interval 0/1 to 2/1, beginning 0/1, 1/4, 1/3, 2/5, 1/2, 3/5, 2/3, 3/4, 1/1, 5/4, 4/3, 7/5, 3/2, 8/5, 5/3, 7/4, 2/1

Farey fractal for interval 0/1 to 3/1, beginning 0/1, 1/3, 1/2, 2/3, 1/1, 5/4, 4/3, 7/5, 3/2, 8/5, 5/3, 7/4, 2/1, 7/3, 5/2, 8/3, 3/1

Farey fractal for interval 0/1 to 4/1, beginning

0/1, 1/3, 1/2, 2/3, 1/1, 4/3, 3/2, 5/3, 2/1, 7/3, 5/2, 8/3, 3/1, 10/3, 7/2, 11/3, 4/1

Farey fractal for interval 0/1 to 5/1, beginning 0/1, 1/1, 5/4, 10/7, 5/3, 7/4, 2/1, 7/3, 5/2, 8/3, 3/1, 13/4, 10/3, 25/7, 15/4, 4/1, 5/1

Farey fractal for interval 0/1 to 6/1, beginning 0/1, 1/2, 1/1, 4/3, 3/2, 5/3, 2/1, 5/2, 3/1, 7/2, 4/1, 13/3, 9/2, 14/3, 5/1, 11/2, 6/1

Farey fractal for interval 0/1 to 7/1, beginning 0/1, 7/5, 7/4, 2/1, 7/3, 21/8, 14/5, 3/1, 7/2, 4/1, 21/5, 35/8, 14/3, 5/1, 21/4, 28/5, 7/1

Farey fractal for interval 0/1 to 8/1, beginning 0/1, 1/2, 1/1, 3/2, 2/1, 5/2, 3/1, 7/2, 4/1, 9/2, 5/1, 11/2, 6/1, 13/2, 7/1, 15/2, 8/1

Farey fractal for interval 0/1 to 9/1, beginning 0/1, 1/1, 3/2, 2/1, 3/1, 7/2, 4/1, 13/3, 9/2, 14/3, 5/1, 11/2, 6/1, 7/1, 15/2, 8/1, 9/1

Farey fractal for interval 0/1 to 10/1, beginning 0/1, 5/4, 5/3, 2/1, 5/2, 3/1, 10/3, 15/4, 5/1, 25/4, 20/3, 7/1, 15/2, 8/1, 25/3, 35/4, 10/1

The shape of the slope is determined by the factorization of *n*:

*n* = 12 = 2^2 * 3

*n* = 16 = 2^4

*n* = 18 = 2 * 3^2

*n* = 20 = 2^2 * 5

*n* = 25 = 5^2

*n* = 27 = 3^3

*n* = 32 = 2^5

*n* = 33 = 3 * 11

*n* = 42 = 2 * 3 * 7

*n* = 64 = 2^6

*n* = 65 = 5 * 13

*n* = 70 = 2 * 5 * 7

*n* = 77 = 7 * 11

*n* = 81 = 3^4

*n* = 96 = 2^5 * 3

*n* = 99 = 3^2 * 11

*n* = 100 = 2^2 * 5^2

Farey fractal-hills, *n* = various

• When Heifetz then played through the work, he made several mistakes in a very difficult passage. The impeccable Heifetz said to Schoenberg: “I’d need to grow six fingers for that!” Schoenberg allegedly replied: “Well, I can wait!”

]]>But you can use other restrictions. For example, suppose that the point can jump only once or twice towards any vertex, that is, (j = 1,2). It can then jump towards the same vertex again, but not the same number of times as it previously jumped. So if it jumps once, it has to jump twice next time; and vice versa. If you use this rule on a pentagon, this fractal appears:

*v* = 5, *j* = 1,2 (black-and-white)

*v* = 5, *j* = 1,2 (colour)

If the point can also jump towards the centre of the pentagon, this fractal appears:

*v* = 5, *j* = 1,2 (with centre)

And if the point can also jump towards the midpoints of the sides:

*v* = 5, *j* = 1,2 (with midpoints)

*v* = 5, *j* = 1,2 (with midpoints and centre)

And here the point can jump 1, 2 or 3 times, but not once in a row, twice in a row or thrice in a row:

*v* = 5, *j* = 1,2,3

*v* = 5, *j* = 1,2,3 (with centre)

Here the point remembers its previous two moves, rather than just its previous move:

*v* = 5, *j* = 1,2,3, *hist* = 2 (black-and-white)

*v* = 5, *j* = 1,2,3, *hist* = 2

*v* = 5, *j* = 1,2,3, *hist* = 2 (with center)

*v* = 5, *j* = 1,2,3, *hist* = 2 (with midpoints)

*v* = 5, *j* = 1,2,3, *hist* = 2 (with midpoints and centre)

And here are hexagons using the same rules:

*v* = 6, *j* = 1,2 (black-and-white)

*v* = 6, *j* = 1,2

*v* = 6, *j* = 1,2 (with centre)

*v* = 8, *j* = 1,2

*v* = 8, *j* = 1,2 (with centre)

*v* = 8, *j* = 1,2,3, *hist* = 2

*v* = 8, *j* = 1,2,3, *hist* = 2

*v* = 8, *j* = 1,2,3,4 *hist* = 3

*v* = 8, *j* = 1,2,3,4 *hist* = 3 (with center)

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